Quod Erat Demonstrandum

2011/03/31

To 6C students

Filed under: Pure Mathematics — johnmayhk @ 5:23 上午

Just a sketch of the solutions to 1-mark questions in the previous quiz.

Q.7

Denote [x] as the greatest integer not greater than x.

Does \displaystyle \lim_{x\rightarrow 0-}(\frac{x}{2011}[\frac{2012}{x}]-\frac{2011}{x}[\frac{x}{2012}]) exist?

Hint:

x is negative \Rightarrow \frac{2012}{x} is negative.

Also, as x tends to 0-, \frac{2012}{x} tends to negative infinity.

Let \frac{2012}{x} = -N+r, where N is a non-negative integer and 0\le r<1

Hence, \frac{x}{2011}[\frac{2012}{x}]=\frac{1}{2011}\frac{2012}{-N+r}[-N+r]=\frac{1}{2011}\frac{2012}{-N+r}(-N)=\frac{2012}{2011}(\frac{1}{1-\frac{r}{N}})

Thus, \displaystyle \lim_{x\rightarrow 0-}\frac{x}{2011}[\frac{2012}{x}] exists.

Now, for x tends to 0-, let \frac{x}{2012}=-s where 0<s<1

\frac{2011}{x}[\frac{x}{2012}]=\frac{2011}{-2012s}[-s]=\frac{2011}{2012s}

Thus, when x tends to 0-, the limit above does not exist.

Q.8

Establish
\pi =2\times\frac{2}{\sqrt{2}}\times\frac{2}{\sqrt{2+\sqrt{2}}}\times\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}\times\dots

Hint:

\sin x
\equiv 2\cos(\frac{x}{2})\sin(\frac{x}{2})
\equiv 2^2\cos(\frac{x}{2})\cos(\frac{x}{2^2})\sin(\frac{x}{2^2})
\dots
\equiv 2^n\cos(\frac{x}{2})\cos(\frac{x}{2^2})\dots \cos(\frac{x}{2^n})\sin(\frac{x}{2^n})

Hence

\frac{\sin x}{2^n\sin(\frac{x}{2^n})}\equiv \cos(\frac{x}{2})\cos(\frac{x}{2^2})\dots \cos(\frac{x}{2^n})

(for \sin(\frac{x}{2^n}))\ne 0)

Taking n tends to infinity, yield

\frac{\sin x}{x}=\cos(\frac{x}{2})\cos(\frac{x}{2^2})\cos(\frac{x}{2^3})\dots

Put x=\frac{\pi}{2}, yield

\frac{2}{\pi}=\cos(\frac{\pi}{2^2})\cos(\frac{\pi}{2^3})\cos(\frac{\pi}{2^4})\dots

Now,

\cos(\frac{\pi}{2^2})=\frac{1}{\sqrt{2}}

\cos(\frac{\pi}{2^3})=\sqrt{\frac{1+\cos(\frac{\pi}{2^2})}{2}}=\sqrt{\frac{1+\frac{1}{\sqrt{2}}}{2}}=\frac{\sqrt{2+\sqrt{2}}}{2}

\cos(\frac{\pi}{2^4})=\sqrt{\frac{1+\frac{\sqrt{2+\sqrt{2}}}{2}}{2}}=\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}

and so on.

3 則迴響 »

  1. 老師您好,請問哪裏有像 Lim 或 dy/dx的練習呢?

    迴響 由 andy wong — 2011/04/01 @ 8:36 下午 | 回覆

    • In web or in my blog? For latter, there should be some in the download page.

      迴響 由 johnmayhk — 2011/04/01 @ 9:54 下午 | 回覆

  2. o…兩個都可以,謝謝

    迴響 由 andy wong — 2011/04/01 @ 10:05 下午 | 回覆


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